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Question: A particle performing UCM changes its angular velocity from \(70\,\,r.p.m\) to \(130\,r.p.m\) in\(18...

A particle performing UCM changes its angular velocity from 70r.p.m70\,\,r.p.m to 130r.p.m130\,r.p.m in18s18\, s. Find the angular acceleration of the particle. (In rads2\dfrac {rad} {{{s} ^ {2}}} )
(A)0.360.36
(B)0.3490.349
(C)0.690.69
(D)0.480.48

Explanation

Solution

This question involves the concept of uniform circular motion. In this question, we will first convert initial and final angular velocities from r.p.m. to r.p.s. then, we will calculate the final and initial angular velocity. After finding this, we will calculate the ratio of the difference between initial angular velocity and final angular velocity to the time taken. Now, we will apply the formula of angular velocity to calculate angular acceleration of the particle.

Complete step by step answer:
We have already been given the angular velocity expression so let us assume,
ω1{{\omega} _ {1}} = initial angular velocity,
ω2{{\omega} _ {2}} = final angular velocity
Now, we will convert the given frequencies from r.p.m. to r.p.s. –
Initial frequency,f1=7060r.p.s.{{f}_{1}}=\,\dfrac{70}{60}r.p.s.
Final frequency,f2=13060r.p.s.{{f}_{2}}=\dfrac{130}{60}r.p.s.
For calculation of angular velocity we use formula,
ω=2πf\omega =2\pi f where, ff = frequency
ω1=2πf1=2π7060=73rads{{\omega} _ {1}} =2\pi {{f} _ {1}} =2\pi \dfrac{70}{60}=\dfrac{7}{3}\dfrac{rad}{s}
ω2=2πf2=2π13060=133rads{{\omega} _ {2}} =2\pi {{f} _ {2}} =2\pi \dfrac{130}{60}=\dfrac{13}{3}\dfrac{rad}{s}
As we know the basic formula of angular velocity so, by applying below expression we get value of angular acceleration:ωt=ω0+αt{{\omega }_{t}}={{\omega }_{0}}+\alpha t where,
ωt{{\omega} _ {t}} = final angular velocity,
ω0{{\omega} _ {0}} =initial angular velocity,
α\alpha =angular acceleration.
\begin {align} & \therefore {{\omega} _ {2}} = {{\omega} _ {1}} +\alpha t \\\ & \Rightarrow \alpha =\dfrac {{{\omega} _ {2}}-{{\omega} _ {1}}} {t} \\\ & \Rightarrow \alpha =\dfrac {\dfrac{13}{3}-\dfrac{7}{3}}{18} \\\ & \Rightarrow \alpha =0.349\dfrac {rad} {{{s} ^ {2}}} \\\ & \\\ \end{aligned}

So, the correct answer is “Option B”.

Note: Angular acceleration is a vector quantity which gives both magnitude and direction. Angular momentum is defined as the vector quantity which is used to measure the rotational momentum of the body or system in rotation. This angular momentum is equal to the product of the angular velocity of the system or body and its moment of inertia with respect to the rotation axis and, this is directed along the rotational axis in classical physics. We must revise the unit of the quantity because we may create mistakes in the unit of angular velocity and angular acceleration.